Question

PQRS and ABRS are parallelogram and x is any point on br show that 1) area(PQRS)=area(ABRS)

2) area(AXS)=1/2 area(PQRS)

Answer 1

Explanation:

Given PQRS and ABRS are parallelogram and X is any point on BR. we have to prove that

1)
`area(PQRS)=area(ABRS)`

2)
`area(AXS)=(1)/(2)area(PQRS)`

In ∆ASP and ΔBRQ

∠SPA = ∠RQB [Corresponding angles]

∠PAS = ∠QBR [Corresponding angles]

PS = QR [Opposite sides of the parallelogram PQRS]

By AAS rule, ∆ASP≅ΔBRQ

∴ ar(ASP) = ar(BRQ) (Congruent triangles have equal area)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ]

= ar(QRB) + ar(ASRQ] = ar(ABRS)

So, ar(PQRS)=ar(ABRS)

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

`ar(AXS)=(1)/(2)ar(ABRS)`

⇒
`ar(AXS)=(1)/(2)ar(PQRS)` (∵ar(PQRS)=ar(ABRS))

Hence Proved.